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\begin{tabular}{l}\( z_{\alpha / 2} \) for the \( 87 \% \) confidence interval \\ \( z_{\alpha / 2}=\square \) \\ \hline\end{tabular}

Ask by Hamilton Mathis. in the United States
Mar 11,2025

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Answer

\( z_{\alpha/2} \approx 1.51 \)

Solution

1. The confidence level is \(87\%\). Therefore, \[ \alpha = 1 - 0.87 = 0.13. \] 2. Since the confidence interval is two‐sided, we split \(\alpha\) into two tails: \[ \frac{\alpha}{2} = \frac{0.13}{2} = 0.065. \] 3. The critical value \(z_{\alpha/2}\) is the positive \(z\) such that the area to its left under the standard normal curve is: \[ 1 - \frac{\alpha}{2} = 1 - 0.065 = 0.935. \] 4. Using a standard normal table or an inverse normal calculator, we find: \[ z_{0.065} \approx 1.51. \] Thus, \[ z_{\alpha/2} \approx 1.51. \]

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To find \( z_{\alpha / 2} \) for an \( 87\% \) confidence interval, you first need to determine \( \alpha \). Since the confidence level is \( 87\% \), \( \alpha = 1 - 0.87 = 0.13 \). Therefore, \( \alpha / 2 = 0.065 \). You can then look up \( z_{0.065} \) in the standard normal distribution table, which gives you approximately \( 1.81 \). Now you have \( z_{\alpha / 2} \approx 1.81 \). This means that about \( 87\% \) of the data lies within the interval defined by \( \mu \pm 1.81\sigma \)! To ensure accuracy, double-check the standard normal table or use a calculator for the precise value, as it gives you a better grasp on how far the estimate extends based on the confidence level!

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